JEE Mathematics 2023 Question with Solution

Given: The normals to the planes are $\hat{i} + \hat{j} + 2\hat{k}$ and $2\hat{i} - \hat{j} + \hat{k}$.

Find: The value of $(\tan^2\theta)(\cot^2\alpha)$.

Using the angle between the normals of the two planes,

Hence,

The angle between a line and a plane is complementary to the angle between the line and the plane's normal. Since the line makes angle $\theta$ with the normal of $P_2$,

Now,

and

Therefore,

Therefore, the required numerical value is $9$.

Given: $P_1: \vec{r} \cdot ( \hat{i} + \hat{j} + 2 \hat{k}) = 9$ and $P_2: \vec{r} \cdot (2 \hat{i} - \hat{j} + \hat{k}) = 15$.

Find: $(\tan^2 \theta)(\cot^2 \alpha)$ where $\theta$ is the angle between the planes and the line makes angle $\theta$ with the normal of $P_2$.

The angle between two planes equals the angle between their normal vectors. So take

Then

Also,

Therefore,

So,

Now the line makes angle $\theta$ with the normal of $P_2$. If $\alpha$ is the angle between the line and the plane $P_2$, then

Thus,

Hence,

So,

Therefore, the answer is $9$.

The second provided approach concludes $81$, but that comes from incorrectly taking $\tan^2\frac{\pi}{3} = 9$ and $\cot^2\frac{\pi}{6} = 9$. The correct values are both $3$, so the correct result remains $9$.